YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) QDPOrderProof [EQUIVALENT, 6 ms]
(4) QDP
(5) MRRProof [EQUIVALENT, 0 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) TRUE


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(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   a(c(x1)) -> c(b(x1))
   a(x1) -> b(b(b(x1)))
   b(c(b(x1))) -> a(c(x1))

Q is empty.

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(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A(c(x1)) -> B(x1)
   A(x1) -> B(b(b(x1)))
   A(x1) -> B(b(x1))
   A(x1) -> B(x1)
   B(c(b(x1))) -> A(c(x1))

The TRS R consists of the following rules:

   a(c(x1)) -> c(b(x1))
   a(x1) -> b(b(b(x1)))
   b(c(b(x1))) -> a(c(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(3) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   A(c(x1)) -> B(x1)
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO]:

   POL(A(x_1)) = x_1
   POL(B(x_1)) = x_1
   POL(a(x_1)) = x_1
   POL(b(x_1)) = x_1
   POL(c(x_1)) = 1 + x_1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   a(x1) -> b(b(b(x1)))
   b(c(b(x1))) -> a(c(x1))
   a(c(x1)) -> c(b(x1))


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(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A(x1) -> B(b(b(x1)))
   A(x1) -> B(b(x1))
   A(x1) -> B(x1)
   B(c(b(x1))) -> A(c(x1))

The TRS R consists of the following rules:

   a(c(x1)) -> c(b(x1))
   a(x1) -> b(b(b(x1)))
   b(c(b(x1))) -> a(c(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(5) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented dependency pairs:

   A(x1) -> B(b(x1))
   A(x1) -> B(x1)
   B(c(b(x1))) -> A(c(x1))

Strictly oriented rules of the TRS R:

   b(c(b(x1))) -> a(c(x1))

Used ordering: Polynomial interpretation [POLO]:

   POL(A(x_1)) = 2 + x_1
   POL(B(x_1)) = x_1
   POL(a(x_1)) = 3 + x_1
   POL(b(x_1)) = 1 + x_1
   POL(c(x_1)) = 3*x_1


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(6)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A(x1) -> B(b(b(x1)))

The TRS R consists of the following rules:

   a(c(x1)) -> c(b(x1))
   a(x1) -> b(b(b(x1)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(7) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
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(8)
TRUE